If i have a vector (array) in matlab, is it possible to remove the values from that vector based on some restriction (e.g.. all non negative numbers).
Can you please advise me on the best approach to do that.
Yes, you can either use logical indexing to keep the values which meet a criterion or use the find function to get the indexes which hold values that meet a criterion.
logical indexing
the find function
An example of logical indexing where we want to remove all the values from a vector which are not greater than three:
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3)
x =
4 5 6
You can also ask for multiple criteria as you would expect. In the following example, we want to keep every value which is greater than three, but not five.
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3 & x~=5)
x =
4 6
Finally, the find function can come in handy when you need the indexes of values which meet a criterion.
>> x=[1,1,2,2,5,5]
x =
1 1 2 2 5 5
>> ind=find(x>3)
ind =
5 6
Logical indexing and find can also be applied to matrices with more than one row/column.
Thanks #Alan for helping me improve the answer.
You may want to look into logical indexing, as it neatly handles your problem.
To use the example you gave, if you have a vector a of numbers, and you want to remove all negative numbers you could do the following:
b = a(a >= 0);
which would create a vector b containing only the positive elements of a, or you could try:
a(a < 0) = [];
would set any elements in the vector a to []
Related
I am familiar with Matlab but am still having trouble with vectorized methods in my intuition, so I was wondering if anyone could demonstrate how they would manage this problem.
I have an array, for example A = [1 1 2 2 1 3 3 3 4 3 4 4 5].
I want to return an array B such that each element is the index of A's most 'recent' element with a different value than the previous ones.
So for our array A, B would equal [x x 2 2 4 5 5 5 8 9 10 10 12], where the x's can be any consistent value you like, because there is no previous index satisfying those characteristics.
I know how I would code it as a for-loop, and I bet the for-loop is probably faster, but can anyone vectorize this to faster than the for-loop?
Here's my for-loop:
prev=0;
B=zeros(length(A),1);
for i=2:length(A)
if A(i-1)~=A(i)
prev=i-1;
end
B(i)=prev;
end
Find the indices of the entries where the value changes:
ind = find(diff(A) ~= 0);
The values that should appear in B are therefore:
val = [0 ind];
Construct the diff of B: fill in the difference between the values that should appear at the right places:
Bd = zeros(size(B))';
Bd(ind + 1) = diff(val);
Now use cumsum to construct B:
B = cumsum(Bd)
Not sure whether this results in a speed-up though.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
I am using Matlab 2012a.
I have an array of k cells (say 1000). I need to find the 5 lowest values of this array and need to do an average of those values in X and Y.
Anyone has an idea how to do that?
Assuming you have arrays X and Y, and you want to find the five lowest Y values:
[m mi] = sort(Y);
lowest5index = mi(1:5);
lowest5Y = Y(lowest5index);
lowest5X = X(lowest5index);
meanYlowest5 = mean(lowest5Y);
meanXlowest5 = mean(lowest5X);
Explanation:
The sort command with two output parameters returns both the sorted array (in m) and the indices in the original array (mi). The first five indices mi(1:5) correspond to the five lowest values. Taking the mean of these values for both X and Y will do what we want. If I didn't understand your problem statement, please clarify your question and I will take another shot at it.
How about doing a sort of your array from lowest value to the highest and then selecting the 5 first values. Those will be the 5 min values of your array. Then perform a mean of those 5 values.
This might not be the most memory efficient way of doing this but for just 1000 values it will get the job done!
Hope it helps!
use minmaxselection MATLAB MEX package, which has been specially optimized for this problem:
a = [2,3,4,7,56,4,21, 64, -2];
mink(a, 2)
<< ans =
<< -2 2
mink(a,4)
<< ans =
<< -2 2 3 4
My question has two parts:
Split a given matrix into its columns
These columns should be stored into an array
eg,
A = [1 3 5
3 5 7
4 5 7
6 8 9]
Now, I know the solution to the first part:
the columns are obtained via
tempCol = A(:,iter), where iter = 1:end
Regarding the second part of the problem, I would like to have (something like this, maybe a different indexing into arraySplit array), but one full column of A should be stored at a single index in splitArray:
arraySplit(1) = A(:,1)
arraySplit(2) = A(:,2)
and so on...
for the example matrix A,
arraySplit(1) should give me [ 1 3 4 6 ]'
arraySplit(2) should give me [ 3 5 5 8 ]'
I am getting the following error, when i try to assign the column vector to my array.
In an assignment A(I) = B, the number of elements in B and I must be the same.
I am doing the allocation and access of arraySplit wrongly, please help me out ...
Really it sounds like A is alread what you want--I can't imagine a scenario where you gain anything by splitting them up. But if you do, then your best bet is likely a cell array, ie.
C = cell(1,3);
for i=1:3
C{i} = A(:,i);
end
Edit: See #EitanT's comment below for a more elegant way to do this. Also accessing the vector uses the same syntax as setting it, e.g. v = C{2}; will put the second column of A into v.
In a Matlab array, each element must have the same type. In most cases, that is a float type. An your example A(:, 1) is a 4 by 1 array. If you assign it to, say, B(:, 2) then B(:, 1) must also be a 4 by 1 array.
One common error that may be biting you is that a 4 by 1 array and a 1 by 4 array are not the same thing. One is a column vector and one is a row vector. Try transposing A(:, 1) to get a 1 by 4 row array.
You could try something like the following:
A = [1 3 5;
3 5 7;
4 5 7;
6 8 9]
arraySplit = zeros(4,1,3);
for i =1:3
arraySplit(:,:,i) = A(:,i);
end
and then call arraySplit(:,:,1) to get the first vector, but that seems to be an unnecessary step, since you can readily do that by accessing the exact same values as A(:,1).
I'm trying to write a script in MATLAB that finds the location of the minimum value of a 2D array of numbers. I am certain there is only 1 minimum in this array, so having multiple locations in the array with the same minimum value is not an issue. I can find the minimum value of the array, but in a 30x30 array, I would like to know which row and column that minimum value is in.
As an alternative version, combine min to get the minimum value and find to return the index, if you've already calculated the minimum then just use find.
>> a=magic(30);
>> [r,c]=find(a==min(min(a)))
r =
1
c =
8
Or depending on how you want to use the location information you may want to define it with a logical array instead, in which case logical addressing can be used to give you a truth table.
>> a=magic(30);
>> locn=(a==min(min(a)));
You could reshape the matrix to a vector, find the index of the minimum using MIN and then convert this linear index into a matrix index:
>> x = randi(5, 5)
x =
5 4 4 2 4
4 2 4 5 5
3 1 3 4 3
3 4 2 5 1
2 4 5 3 5
>> [value, index] = min(reshape(x, numel(x), 1));
>> [i,j] = ind2sub(size(x), index)
i =
3
j =
2
Look at the description of the min function. It can return the minimum value as well as the index. For a two dimensional array, just call it twice.
A = rand(30); % some matrix
[minColVal, minColIdx] = min(A);
[minRowVal, minRowIdx] = min(minColVal);
minVal = minRowVal;
minValIdx = [minColIdx(minRowIdx), minRowIdx];
Edit: #b3's solution is probably computationally more elegant (faster and needs less temporary space)
To find min or max in a subset of a vector -
If A is a vector and "lowerBound" and "upperBound" are the bounds of the vector among which you need to find the max (or min) value, then use this command -
[Value,Index]=min(A(lowerBound:upperBound));
This returns "Value" as the min or max value among A(lowerBound) and A(uppedBound) and
"Index" as with "lowerBound" as the offset. So to find the absolute index, you need to add "lowerBound" to the Index.
An alternate solution using an inline function will work.
>> min_index = #(matrix) find(matrix == min(reshape(matrix, [1,numel(matrix)])));
>> a=magic(30);
>> [r,c]=min_index(a)
r =
1
c =
8